Rescale $x$, then rescale $t$ identically so the derivatives both get the same constant multipliers. The constants that pop out don't affect the signum, so you can rescale $\varphi$. So this is a kind of scaling limit of sinh-Gordon.
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Steve HuntsmanApr 25 '10 at 16:20

2

I thought about that but sinh would scale like O(\phi) in the small limit while sgn(\phi) scales like O(1). Thanks for the idea of scaling though.
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Kaveh KhodjastehApr 25 '10 at 17:10

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Jim, Kaveh is talking about the sgn function, not the sin function. (That is, it's a pun on the sine-Gordon equation.)
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Qiaochu YuanApr 25 '10 at 18:28

Sorry, I was distracted by the header and didn't look far enough into the question. Some of my colleagues have been fond of sine-Gordon, but I'm an outsider.
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Jim HumphreysApr 25 '10 at 19:23